string 从 Delphi 字符串中删除 '#$A'
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Removing '#$A' from Delphi string
提问by JCTLK
I am modifying a delphi app.In it I'm getting a text from a combo box. The problem is that when I save the text in the table, it contains a carriage return. In debug mode it shows like this.
我正在修改一个 delphi 应用程序。在它里面我从一个组合框中获取一个文本。问题是当我将文本保存在表格中时，它包含一个回车符。在调试模式下，它显示如下。
newStr := 'Projector Ex320u-st Short Throw '#$A'1024 X 768 2700lm'
Then I have put
然后我把
newStr := StringReplace(newStr,'#$A','',[rfReplaceAll]);
to remove the '#$A' thing. But this doesn't remove it.
删除 '#$A' 的东西。但这并不能消除它。
Is there any other way to do this..
有没有其他方法可以做到这一点..
Thanks
谢谢
回答by Marjan Venema
Remove the quotes around the #$A:
删除#$A 周围的引号：
newStr := StringReplace(newStr,#$A,'',[rfReplaceAll]);
The # tells delphi that you are specifying a character by its numerical code. The $ says you are specifying in Hexadecimal. The A is the value.
# 告诉 delphi 您正在通过其数字代码指定一个字符。$ 表示您以十六进制指定。A 是值。
With the quotes you are searching for the presence of the #$A characters in the string, which aren't found, so nothing is replaced.
使用引号，您正在搜索字符串中是否存在 #$A 字符，但未找到这些字符，因此不会替换任何内容。
回答by jcfaria
Adapted from http://www.delphipages.com/forum/showthread.php?t=195756
改编自http://www.delphipages.com/forum/showthread.php?t=195756
The '#' denotes an ASCII character followed by a byte value (0..255).
'#' 表示一个 ASCII 字符后跟一个字节值 (0..255)。
The $Ais hexadecimal which equals 10and $Dis hexadecimal which equals 13.
的$A是十六进制，等于10和$D十六进制，等于13。
#$Aand #$D(or #10and #13) are ASCII line feed and carriage return characters respectively.
#$Aand #$D(或#10and #13) 分别是 ASCII 换行符和回车符。
Line feed = ASCII character $A(hex) or 10(dec): #$Aor #10
换行符 = ASCII 字符$A（十六进制）或10（十进制）：#$A或#10
Carriage return = ASCII character $D(hex) or 13(dec): #$Dor #13
回车符 = ASCII 字符$D（十六进制）或13（十进制）：#$D或#13
So if you wanted to add 'Ok' and another line:
因此，如果您想添加“确定”和另一行：
Memo.Lines.Add('Ok' + #13#10)
or
或者
Memo.Lines.Add('Ok' + #$D#$A)
To remove the control characters (and white spaces) from the beginning and end of a string:
从字符串的开头和结尾删除控制字符（和空格）：
MyString := Trim(MyString)
Why doesn't Pos() find them?
为什么 Pos() 找不到它们？
That is how Delphi displays control characters
to you, if you were to do Pos(#13, MyString)or Pos(#10, MyString)then it
would return the position.
这就是 Delphi 向您显示控制字符的方式，如果您要这样做，Pos(#13, MyString)否则Pos(#10, MyString)它将返回位置。
